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Topic: Solving e^y - 3e^(-y) + 2 = 0
Replies: 2   Last Post: Jan 30, 2006 5:30 AM

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ticbol

Posts: 116
Registered: 1/25/05
Re: Solving e^y - 3e^(-y) + 2 = 0
Posted: Jan 30, 2006 5:30 AM
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- wrote:
> Please assist. I'm stuck with this:
>
> e^y - 3e^(-y) + 2 = 0
> e^y - 3(e^y)^-1 + 2 = 0
> Let x = e^y
> x - 3x^-1 + 2 = 0
> I don't know what to do next (and am unsure if the above is correct)


---------------------------------------------
Here is one way.

e^y -3e^(-y) +2 = 0
e^y -3/(e^y) +2 = 0
Clear the fraction, multiply both sides by e^y,
(e^y)^2 -3 +2e^y = 0
(e^y)^2 +2e^y -3 = 0
(e^y +3)(e^y -1) = 0

e^y +3 = 0
e^y = -3
Take the natural log of both sides,
y*ln(e) = ln(-3)
y = ln(-3) ----------------cannot be, so reject this.

e^y -1 = 0
e^y = 1
y*ln(e) = ln(1)
y = 0 ------------------------answer.

Check,
e^y -3e^(-y) +2 = 0
e^0 -3e^(-0) +2 =? 0
1 -3(1) +2 =? 0
1 -3 +2 =? 0
0 = 0
Yes, so, OK.

----------------
Is negative zero, zero?
If you do not like that, then -3e^(-y) = -3/(e^y) -----> -3/(e^0) =
-3/1 = -3.




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